माना कि ABC एक त्रिभुज है जिसमें `/_ACB =(pi)/(6)` तथा `A,B ` तथा `C ` की सम्मुख भुजाओं की लम्बाइयाँ क्रमशः a,b तथा C है। यदि `a=x^(2) + x +1 , b=x^(2)-1` तथा ` c=2x+1` हो तब x का मान ज्ञात करों।

Explanation

हम जानते हैं।
`cos C = (a^(2) + b^(2) –c^(2))/(2ab)`
जहाँ `C=(pi)/(6), a=x^(2) +x+ 1,b=x^(2)-1`
तथा` c= 2x+1` रखने पर
`cos .(pi)/(6) = ((x^(2) +x+1)^(2) +(x^(2) -1)^(2)-(2x+1)^(2))/(2(x^(2)+x+1)(x^(2)-1))`
`sqrt(3) =((x^(2)-1)^(2)+(x^(2)+x+1+2x+1)(x^(2)+x+1-2x-1))/((x^(2)+x+1)(x^(2)-1))`
`sqrt(3) =((x^(2)-1)^(2)+[x(x+1)+2(x+1)](x^(2)-x))/((x^(2)+x+1)(x^(2)-1))`
`sqrt(3)= ((x^(2) -1)^(2)+(x+1)(x+2)x(x-1))/((x^(2)+x+1)(x^(2)-1)) =((x^(2)-1)^(2)+x(x^(2)-1)(x+2))/((x^(2)+x+1)(x^(2)-1))`
`sqrt(3) = ((x^(2)-1)[x^(2)-1+x^(2)+2x])/((x^(2) +x+1)(x^(2)-1))`
`sqrt(3) = (2x^(2) +2x+1)/(x^(2)+x+1)`
`sqrt(3)x^(2) +sqrt(3x) +sqrt(3)=2x^(2) +2x+1`
`(sqrt(3)-2)x^(2) +(sqrt(3)-2) x+ (sqrt(3)+1) = 0`
श्रीधराचार्य सूत्रानुसार ,
`=(-(sqrt(3)-2)+-sqrt((7-4sqrt(3)-(4-4sqrt(3)))))/(2(sqrt(3)-2))`
`=(-(sqrt(3)-2)+-sqrt(3))/(2(sqrt(3)-2))`
`=(-sqrt(3)+2+sqrt(3))/(2(sqrt(3)-2))`या `(-sqrt(3)+2-sqrt(3))/(2sqrt(3)-2)`
`=(1)/(sqrt(3)-2)` या `(1-sqrt(3))/(sqrt(3)-2)`
`=-(sqrt(3)+2)`या ` sqrt(3)+1`